Question

If $I_{1}=\displaystyle \int _{0}^{\pi }\frac{x \text{sin} x}{1 + \text{cos}^{2} x}dx\text{, }I_{2}=\displaystyle \int _{0}^{\pi }x\text{sin}^{4}xdx$ then, $I_{1}\text{:}I_{2}$ is equal to

• A. $3 : 4$
• B. $1 : 2$
• C. $4 : 3$
• D. $2 : 3$

Answer 1

Answer: (C)

Given, $I_{1} = \displaystyle \int _{0}^{\pi } \frac{\left(\pi - x\right) \text{sin} \left(\pi - x\right)}{1 + \left(\text{cos} \left(\pi - x\right)\right)^{2}} d x \, \, \, \left(\displaystyle \int _{0}^{a} f \left(x\right) d x = \displaystyle \int _{0}^{a} f \left(a - x\right) d x\right)$
$= \pi \displaystyle \int _{0}^{\pi } \frac{sin x}{1 + \text{cos}^{2} x} d x - \displaystyle \int _{0}^{\pi } \frac{x sin x}{1 + \text{cos}^{2} x} d x$
$2 I_{1} = \pi \displaystyle \int _{0}^{\pi } \frac{sin x}{1 + \text{cos}^{2} x} d x = 2 \pi \displaystyle \int _{0}^{\frac{\pi }{2}} \frac{sin x}{1 + \text{cos}^{2} x} d x$
$\Rightarrow \, I_{1} = \pi \displaystyle \int _{0}^{\frac{\pi }{2}} \frac{sin x}{1 + \left(\text{cos}\right)^{2} x} d x = \pi \displaystyle \int _{0}^{1} \frac{d t}{1 + t^{2}} \, \left(t = cos x\right)$
$= \left[\pi \text{tan}^{-1} t\right]_{0}^{1} = \frac{\pi ^{2}}{4}$
$I_{2}=\displaystyle \int _{0}^{\pi }\left(\pi - x\right)\left(\text{sin}\right)^{4}xdx$
$=\pi \displaystyle \int _{0}^{\pi }\text{sin}^{4}xdx-I_{2}$
$\Rightarrow \, 2I_{2}=2\pi \displaystyle \int _{0}^{\pi / 2}\text{sin}^{4}xdx=2\pi \cdot \frac{3}{4}\cdot \frac{1}{2}\cdot \frac{\pi }{2}$
$\Rightarrow I_{2}=\frac{3}{16}\pi ^{2}$
Therefore, $I _{1} : I ⁡_{2} = \frac{1}{4} : \frac{3}{16} = 4 : 3$